AtCoder Regular Contest 153 E Deque Minimization

// Problem: E - Deque Minimization
// Contest: AtCoder - AtCoder Regular Contest 153
// URL: https://atcoder.jp/contests/arc153/tasks/arc153_e
// Memory Limit: 1024 MB
// Time Limit: 5000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

/*
考虑先写出区间 dp 转移式子。

考虑有转移则连边，抽象到在网格图上走，从主对角线到 $(1, n)$ 方案数。

删除无用转移，转化为 $|\Sigma|$ 次在开头加入若干个 $1$，做若干次前缀和。次数取决于段的长度。

NTT 优化即可。
*/

const int maxn = 1000100;
const int N = 1000000;
const ll mod = 998244353, G = 3;

inline ll qpow(ll b, ll p) {
	ll res = 1;
	while (p) {
		if (p & 1) {
			res = res * b % mod;
		}
		b = b * b % mod;
		p >>= 1;
	}
	return res;
}

ll n, fac[maxn], ifac[maxn], f[maxn], g[maxn];
char s[maxn];

struct node {
	int l, r, x;
	node(int a = 0, int b = 0, int c = 0) : l(a), r(b), x(c) {}
} a[maxn];

inline void init() {
	fac[0] = 1;
	for (int i = 1; i <= N; ++i) {
		fac[i] = fac[i - 1] * i % mod;
	}
	ifac[N] = qpow(fac[N], mod - 2);
	for (int i = N - 1; ~i; --i) {
		ifac[i] = ifac[i + 1] * (i + 1) % mod;
	}
}

inline ll C(ll n, ll m) {
	if (n < m || n < 0 || m < 0) {
		return 0;
	} else {
		return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
	}
}

int r[maxn];

typedef vector<ll> poly;

inline poly NTT(poly a, int op) {
	int n = (int)a.size();
	for (int i = 0; i < n; ++i) {
		if (i < r[i]) {
			swap(a[i], a[r[i]]);
		}
	}
	for (int k = 1; k < n; k <<= 1) {
		ll wn = qpow(op == 1 ? G : qpow(G, mod - 2), (mod - 1) / (k << 1));
		for (int i = 0; i < n; i += (k << 1)) {
			ll w = 1;
			for (int j = 0; j < k; ++j, w = w * wn % mod) {
				ll x = a[i + j], y = w * a[i + j + k] % mod;
				a[i + j] = (x + y) % mod;
				a[i + j + k] = (x - y + mod) % mod;
			}
		}
	}
	if (op == -1) {
		ll inv = qpow(n, mod - 2);
		for (int i = 0; i < n; ++i) {
			a[i] = a[i] * inv % mod;
		}
	}
	return a;
}

inline poly operator * (poly a, poly b) {
	a = NTT(a, 1);
	b = NTT(b, 1);
	int n = (int)a.size();
	for (int i = 0; i < n; ++i) {
		a[i] = a[i] * b[i] % mod;
	}
	a = NTT(a, -1);
	return a;
}

inline void calc(int l, int r, int t) {
	int n = r - l, k = 0;
	while ((1 << k) <= (n + 1) * 2) {
		++k;
	}
	for (int i = 1; i < (1 << k); ++i) {
		::r[i] = (::r[i >> 1] >> 1) | ((i & 1) << (k - 1));
	}
	poly A(1 << k), B(1 << k);
	for (int i = 0; i <= n; ++i) {
		A[i] = f[l + i];
		B[i] = C(i + t - 1, t - 1);
	}
	poly res = A * B;
	for (int i = 0; i <= n; ++i) {
		f[l + i] = res[i];
	}
}

void solve() {
	scanf("%s", s + 1);
	n = strlen(s + 1);
	int m = 0;
	for (int i = 1, j = 1; i <= n; i = (++j)) {
		while (j < n && s[j + 1] == s[i]) {
			++j;
		}
		if (m && s[i] - '0' < a[m].x) {
			break;
		}
		a[++m] = node(i, j, s[i] - '0');
	}
	for (int i = m; i; --i) {
		int j = a[i].r;
		while (j < n && s[j + 1] - '0' > a[i].x) {
			++j;
		}
		for (int k = a[i].l; k <= a[i].r; ++k) {
			f[k] = 1;
		}
		calc(a[i].r, j, a[i].r - a[i].l + 1);
	}
	printf("%lld\n", f[n]);
}

int main() {
	init();
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}